A Question on Sequence of Measurable Functions Given that $(X,\mathcal{F},\mu)$ is a finite measure space and $\{f_k\}$ is a sequence of finite-valued measurable functions, such that for any $\varepsilon > 0$
$$ \lim_{n \rightarrow \infty} \mu(\{x \in X : \sup_{k\geq n} |f_k(x)| \geq \varepsilon\}) =0 .$$
I need to show that $\lim_{k \rightarrow \infty} f_k(x) = 0$ $\mu$-a.e on $X$.
The equality is easy to show. But I am stuck at showing explicitly that the limit $\lim_{k \rightarrow \infty} f_k(x)$ exists $\mu$-a.e on $X$.
Any help/comment is greatly appreciated !
Note that by "finite-valued function", it is only meant that the function does not take the values $+\infty$ or $-\infty$ on $X$.
 A: Fix $\varepsilon>0$. Let $g_n=\sup\{|f_k|:\ k\geq n\}$. Our assumption is that $\lim_n\mu\{g_n\geq\varepsilon\}=0$. We can write this as $\lim_n\mu\{g_n<\varepsilon\}=\mu(X)$.
The sequence $\{g_n\}$ is decreasing and positive; in particular it is convergent. So the sets $\{g_n<\varepsilon\}$ form an increasing sequence. Then, using continuity of the measure, 
$$
\mu(\bigcup_n\{g_n<\varepsilon\})=\lim_n\mu(\{g_n<\varepsilon\})=\mu(X).
$$
That is, up to a null-set, $X=\bigcup_n\{g_n<\varepsilon\}$. In other words, $\lim_ng_n<\varepsilon$ a.e. As $\varepsilon$ was arbitrary, $\lim_ng_n=0$. This means that 
$$
\limsup_k|f_k|=0\ \text{ a.e. },
$$
which implies $\lim_k|f_k|=0$ a.e. 
A: Let $$A_{n,\epsilon} = \{|f_n| \geq \epsilon \}, $$
$$A_\epsilon  = \limsup_n A_{n,\epsilon} = \bigcap_{n=1}^\infty\bigcup_{k=n}^{\infty}A_{k,\epsilon}.$$
As $\bigcup_{k=n}^{\infty}A_{k,\epsilon}$ is decreasing with intersection over $n$ equal to $ A_\epsilon $, we have:
$$\lim_n \mu\left(\bigcup_{k=n}^{\infty}A_{k,\epsilon} \right) = \mu\left(A_\epsilon \right).  $$
Now:
$$ \left\{\lim_n f_n \not= 0\right\}  = \bigcup_{\epsilon>0} A_\epsilon =\bigcup_{m=1}^{\infty} A_{1/m}.$$ 
Last equality is needed to show the measurability of this set.
So:
$\lim_n f_n = 0$ a.e. if and only if $$\mu\left(A_\epsilon \right)=0$$ for all $\epsilon > 0$ if and only if $$\lim_n \mu\left(\bigcup_{k=n}^{\infty}A_{k,\epsilon} \right) =0$$ for all   $\epsilon > 0$. 
Finally, note that
$$ \bigcup_{k=n}^{\infty}A_{k,\epsilon}  \subseteq \left\{\sup_{k\geq n} |f_k|\geq \epsilon\right\}.$$
